物理代写|统计力学代写Statistical mechanics代考|PHYS3006

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

物理代写|统计力学代写Statistical mechanics代考|Multiplying probabilities

How is $P(A \cap B)$ in Eq. (3.2) calculated? To answer that, it’s necessary to introduce another kind of probability, the conditional probability, denoted $P(A \mid B)$, the probability of $A$ occurring, given that $B$ has occurred. Referring to Fig. 3.3, we’re interested in the probability that $A$ occurs given that $B$ has definitely occurred, a type of problem where the sample space has changed-in this case the certain event is $B$. The probability we want is the ratio of the number of sample points in the intersection, $N_{A \cap B}$, to that in $B$ :
$$P(A \mid B)=\frac{\bar{N}{A \cap B}}{N{B}}=\frac{\bar{N}{A \cap B}}{N{\Omega}} \frac{\bar{N}{\Omega}}{N{B}}=\frac{\bar{P}(A \cap \bar{B})}{P(B)},$$
or
$$P(A \wedge B)=P(A \mid B) P(B) .$$
In words, Eq. (3.4) indicates that the probability of $A$ and $B$ is the probability of $A$ given that $B$ has occurred, multiplied by the probability that $B$ occurs. This relation is symmetrical between $A$ and $B: P(A \cap B)=P(B \mid A) P(A)$, implying $P(B \mid A)=P(A \mid B) P(B) / P(A)$

Suppose $A$ and $B$ are such that $P(A \mid B)=P(A)$. In that case $A$ is said to be independent of $B$-the probability of $A$ occurring is independent of the condition that $B$ has occurred. For independent events, Eq. (3.4) reduces to
$$P(A \cap B)=P(A) P(B) . \quad \text { (independent events) }$$
For independent events, the probability of $A$ and $B$ is the product of the probabilities. Many problems in physics implicitly assume independent events; many problems implicitly ask for the probability of “this and that and that.” Be on the lookout for how statements are worded; there may be implied “ands.” Thus, for mutually exclusive events, probabilities are added, Eq. (3.3), whereas for independent events, probabilities are multiplied, Eq. (3.5). In Section 3.4, we give examples of how to calculate probabilities using these rules. First we must learn to count.

物理代写|统计力学代写Statistical mechanics代考|Stirling’s approximation

In its simplest form, Stirling’s approximation is, for $n \gg 1$,
$$\ln n !=n(\ln n-1)+O(\ln n),$$
where $O(\ln n)$ indicates that terms of order $\ln n$ have been neglected (which are negligible compared to $n$ for large $n$ ). Equation (3.14) is one of those results that should work only for $n \rightarrow \infty$, but which is accurate for relatively small values of $n(n \approx 10)$; see Exercise 3.8. Equation (3.14) is surprisingly easy to derive: $\ln n !=\sum_{k=1}^{n} \ln k \approx \int_{1}^{n} \ln x \mathrm{~d} x=\left.(x \ln x-x)\right|{1} ^{n} \approx n \ln -n$. The $O(\ln n)$ remainder is evaluated below. A more accurate version of Stirling’s approximation is $$n !^{n \rightarrow \infty} \underset{\sim}{2 \pi n}\left(\frac{n}{\mathrm{e}}\right)^{n},$$ where the notation $\sim$ indicates asymptotic equivalence. ${ }^{8}$ Equation (3.15) can be derived from $\Gamma(x)$ (see Eq. (B.1)): $$\Gamma(n+1)=n !=\int{0}^{\infty} x^{n} \mathrm{e}^{-x} \mathrm{~d} x \stackrel{x=n y}{=} n n^{n} \int_{0}^{\infty} \mathrm{e}^{n(\ln y-y)} \mathrm{d} y .$$
The integral on the right of Eq. (3.16) can be approximated using the method of steepest descent[16. p233] for large $n$ :
$$\int_{0}^{\infty} \mathrm{e}^{n(\ln y-y)} \mathrm{d} y \sim \sqrt{\frac{2 \pi}{n}} \mathrm{e}^{-n} .$$
Combining Eqs. (3.17) and (3.16) yields Eq. (3.15). By taking the logarithm of Eq. (3.15), we see that the remainder term in Eq. (3.14) is $\frac{1}{2} \ln (2 \pi n)$.

Sometimes we require the logarithm of the gamma function (the log-gamma function), $\ln \Gamma(x)$. From the recursion relation, Eq. (B.3), $\ln \Gamma(x+1)=\ln x+\ln \Gamma(x)$, and thus
$$\ln \Gamma(x)=\ln \Gamma(x+1)-\ln x .$$
Use Stirling’s approximation,
$$\Gamma(x+1) \sim \sqrt{2 \pi x}\left(\frac{x}{\mathrm{e}}\right)^{x} .$$

物理代写|统计力学代写Statistical mechanics代考|Multiplying probabilities

$$P(A \mid B)=\frac{\bar{N} A \cap B}{N B}=\frac{\bar{N} A \cap B}{N \Omega} \frac{\bar{N} \Omega}{N B}=\frac{\bar{P}(A \cap \bar{B})}{P(B)},$$

$$P(A \wedge B)=P(A \mid B) P(B) .$$

$$P(A \cap B)=P(A) P(B) . \quad \text { (independent events) }$$

物理代写|统计力学代写Statistical mechanics代考|Stirling’s approximation

$$\ln n !=n(\ln n-1)+O(\ln n),$$

$\ln n !=\sum_{k=1}^{n} \ln k \approx \int_{1}^{n} \ln x \mathrm{~d} x=(x \ln x-x) \mid 1^{n} \approx n \ln -n$. 这 $O(\ln n)$ 余数在下面评估。斯特林近似的 更准确版本是
$$n !^{n \rightarrow \infty} \underset{\sim}{2 \pi n}\left(\frac{n}{\mathrm{e}}\right)^{n},$$

$$\Gamma(n+1)=n !=\int 0^{\infty} x^{n} \mathrm{e}^{-x} \mathrm{~d} x \stackrel{x=n y}{=} n n^{n} \int_{0}^{\infty} \mathrm{e}^{n(\ln y-y)} \mathrm{d} y .$$

$$\int_{0}^{\infty} \mathrm{e}^{n(\ln y-y)} \mathrm{d} y \sim \sqrt{\frac{2 \pi}{n}} \mathrm{e}^{-n} .$$

$$\ln \Gamma(x)=\ln \Gamma(x+1)-\ln x .$$

$$\Gamma(x+1) \sim \sqrt{2 \pi x}\left(\frac{x}{\mathrm{e}}\right)^{x} .$$

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MATLAB代写

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